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On 08/05/2012 05:04 PM, Jim Henderson wrote:
> The error is when you've reached the end of the trail and keep going.
If you set out to fail, and succeed, what have you done?
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On Tue, 08 May 2012 19:05:04 +0100, Orchid Win7 v1 wrote:
> On 08/05/2012 05:04 PM, Jim Henderson wrote:
>
>> The error is when you've reached the end of the trail and keep going.
>
> If you set out to fail, and succeed, what have you done?
You've taken a ride down the fail trail? ;)
Jim
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>> If you set out to fail, and succeed, what have you done?
>
> You've taken a ride down the fail trail? ;)
Does the set of all sets that do not contain their own complement
contain itself?
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On 5/8/2012 12:44 PM, Orchid Win7 v1 wrote:
> Does the set of all sets that do not contain their own complement
> contain itself?
Yes
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On 08/05/2012 09:02 PM, Kevin Wampler wrote:
> On 5/8/2012 12:44 PM, Orchid Win7 v1 wrote:
>> Does the set of all sets that do not contain their own complement
>> contain itself?
>
> Yes
That's great. Now tell me, is the set of all sets that list themselves
listed in itself?
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On Tue, 08 May 2012 20:44:36 +0100, Orchid Win7 v1 wrote:
>>> If you set out to fail, and succeed, what have you done?
>>
>> You've taken a ride down the fail trail? ;)
>
> Does the set of all sets that do not contain their own complement
> contain itself?
You do realise that I'm razzing you about your spelling again, right? ;)
Jim
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On 5/8/2012 1:23 PM, Orchid Win7 v1 wrote:
> On 08/05/2012 09:02 PM, Kevin Wampler wrote:
>> On 5/8/2012 12:44 PM, Orchid Win7 v1 wrote:
>>> Does the set of all sets that do not contain their own complement
>>> contain itself?
>>
>> Yes
>
> That's great. Now tell me, is the set of all sets that list themselves
> listed in itself?
>
I assume by "list" you mean "contain"? If so, then I think the answer
is no.
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>>>> If you set out to fail, and succeed, what have you done?
>>>
>>> You've taken a ride down the fail trail? ;)
>>
>> Does the set of all sets that do not contain their own complement
>> contain itself?
>
> You do realise that I'm razzing you about your spelling again, right? ;)
Yeah, I had noticed. :-P
I was hoping to blind you with library science...
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>> That's great. Now tell me, is the set of all sets that list themselves
>> listed in itself?
>
> I assume by "list" you mean "contain"?
Yes.
> If so, then I think the answer is no.
And I think the answer is "Russell's paradox".
Interestingly, in Portal 2 GLaDDOS poses nearly the exact same question
to Wheatly, who also responds with "false".
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On 5/8/2012 1:52 PM, Orchid Win7 v1 wrote:
> And I think the answer is "Russell's paradox".
That answer is also correct. I'm being overly pedantic in my answers
but here's my reasoning:
----
1) Does the set of all sets that do not contain their own complement
contain itself?
There is no such set, and thus the statement is vacuously true.
2) Now tell me, is the set of all sets that list themselves listed in
itself?
There are no sets which contain themselves, thus the set of all these
sets it in fact the empty set. The empty set does not contain itself,
so the answer is "no".
----
Of course both of these answers are assuming modern "post Russel's
paradox" set theory, but I think they work.
> Interestingly, in Portal 2 GLaDDOS poses nearly the exact same question
> to Wheatly, who also responds with "false".
Maybe an in-joke by some math-savvy writers? Assuming my reasoning is
correct that is.
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